Вступ до аналізу. Ч. 1
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2). α = 2; y = x2 ; D |
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= (− ∞, + ∞), E |
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= [0, + ∞) . 16. |
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3) |
α = 3; y = x3 ; D |
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= E |
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= (− ∞, + ∞). , ! - |
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!:
$ . 19.
4) |
α = |
1 ; y = x 2 |
= x ; D = E = [0, + ∞). |
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$ . 20. |
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5) α = −1; y = x−1 = |
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; D |
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= E |
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= (− ∞;0) (0; + ∞). - |
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$ . 21.
6) |
α = −2; y = x−2 = |
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; D |
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= (− ∞; + 0) (0; + ∞) , E |
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= (0; + ∞). |
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2 |
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$ . 22.
% : y = a x (a > 0, a ≠ 1). Df = (−∞; + ∞), E f = (0; + ∞) A-
+ * a :
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1) a > 1 |
2) 0 < a <1 |
$ . 23 ( ). $ . 23 (!).
6, ( + * ( , " +
", * 2 ! ( " *2.
4. 8 : |
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y = loga x (a > 0, a ≠ 1); D f = (0; + ∞), E f |
= (−∞; + ∞) . A + |
+ * a : |
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1) a > 1 |
2) 0 < a <1 |
$ . 24 ( ). |
$ . 24 (!). |
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5. # :
1) y = sin x; D f = (− ∞; + ∞), E f = [−1; 1].
$ . 25 ( ).
2) y = cos x ; D f = (− ∞ , + ∞), E f = [−1;1] .
$ . 25 (!).
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π |
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E f = (− ∞; + ∞) ( . 26( )); |
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3) |
y = tg x ; Df |
= / x = |
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+ πk , k , |
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y = ctg x ; D f = /{x = πk , k }, E f |
= (− ∞, + ∞) ( . 26(!)) |
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$ . 26 ( ). |
$ . 26 (!). |
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6. )! .
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= [−1;1], |
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π |
π |
( . 27( )) |
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1) y = arcsin x ; Df |
E f |
= |
− |
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y = arccos x ; D f |
= [−1;1], |
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= [0; π] |
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( . 27(!) |
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$ . 27 ( ).
2) y = arctg x; D f = (− ∞, + ∞), E f
y = arcctg x ; D f = (− ∞; + ∞), E f
$ . 27 (!).
=− π , π ( . 28( ))
2 2
=(0; π) ( . 28(!))
$ . 28 ( ). |
$ . 28 (!). |
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